A shell and tube heat exchanger with one shell pass

Asked by admin @ 27/11/2021 in Engineering viewed by 424 People

A shell-and tube heat exchanger (two shells, four tube passes) is used to heat 10,000 kg/h of pressurized water from 35 to 120 oC with 5000 kg/h pressurized water entering the exchanger at 300 oC. If the overall heat transfer coefficient is 1500 W/m^2-K, determine the required heat exchanger area.

Answered by admin @ 27/11/2021

Answer:

4.75m^2

Explanation:

Given:-

- Temperature of hot fluid at inlet: $T_h_i = 300$ °C

- Temperature of cold fluid at outlet: $T_c_o = 120$ °C

- Temperature of cold fluid at inlet: $T_c_i = 35$ °C

- The overall heat transfer coefficient: U = 1500 W / m^2 K

- The flow rate of cold fluid: m_c = 10,00 kg/ h

- The flow rate of hot fluid: m_h = 5,000 kg/h

Solution:-

- We will evaluate water properties at median temperatures of each fluid using table A-4.

Cold fluid: Tci = 35°C , Tco = 35°C

Tcm = 77.5 °C ≈ 350 K --- > $C_p_c = 4195 \frac{J}{kg.K}$

Hot fluid: Thi = 300°C , Tho = 150°C ( assumed )

Thm = 225 °C ≈ 500 K --- > $C_p_h = 4660 \frac{J}{kg.K}$

- We will use logarithmic - mean temperature rate equation as follows:

$A_s = \frac{q}{U*dT_l_m}$

Where,

A_s : The surface area of heat exchange

ΔT_lm: the logarithmic differential mean temperature

q: The rate of heat transfer

- Apply the energy balance on cold fluid as follows:

$q = m_c * C_p_c * ( T_c_o - T_c_i )\\\\q = \frac{10,000}{3600} * 4195 * ( 120 - 35 )\\\\q = 9.905*10^5 W$

- Similarly, apply the heat balance on hot fluid and evaluate the outlet temperature ( Tho ) :

$T_h_o = T_h_i - \frac{q}{m_h * C_p_h} \\\\T_h_o = 300 - \frac{9.905*10^5}{\frac{5000}{3600} * 4660} \\\\T_h_o = 147 C$

- We will use the experimental results of counter flow ( unmixed - unmixed ) plotted as figure ( Fig . 11.11 ) of the " The fundamentals to heat transfer" and determine the value of ( P , R , F ).

- So the relations from the figure 11.11 are:

$P = \frac{T_c_o - T_c_i}{T_h_i - T_c_i} \\\\P = \frac{120 - 35}{300 - 35} \\\\P = 0.32$

$R = \frac{T_h_i - T_h_o}{T_c_o - T_c_i} \\\\R = \frac{300 - 147}{120 - 35} \\\\R = 1.8$

Therefore, P = 0.32 , R = 1.8 ---- > F ≈ 0.97

- The log-mean temperature ( ΔT_lm - cf ) for counter-flow heat exchange can be determined from the relation:

$dT_l_m = \frac{( T_h_i - T_c_o ) - ( T_h_o - T_c_i ) }{Ln ( \frac{( T_h_i - T_c_o )}{( T_h_o - T_c_i )} ) } \\\\dT_l_m = \frac{( 300 - 120 ) - ( 147 - 35 ) }{Ln ( \frac{( 300-120 )}{( 147-35)} ) } \\\\dT_l_m = 143.3 K$

- The log - mean differential temperature for counter flow is multiplied by the factor of ( F ) to get the standardized value of log - mean differential temperature:

$dT_l = F*dT_l_m = 0.97*143.3 = 139 K$

- The required heat exchange area ( A_s ) can now be calculated:

$A_s = \frac{9.905*10^5 }{1500*139} \\\\A_s = 4.75 m^2$

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